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Proper Refinement of Datalog Clauses using Primary Keys

Proper Refinement of Datalog Clauses using Primary Keys. Siegfried Nijssen and Joost N. Kok BNAIC-2003, Nijmegen. Introduction. Inductive Logic Programming algorithm:. C : Set of Datalog clauses, initially empty D : Database of facts (Knowledge base) repeat

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Proper Refinement of Datalog Clauses using Primary Keys

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  1. Proper Refinement of Datalog Clauses using Primary Keys Siegfried Nijssen and Joost N. Kok BNAIC-2003, Nijmegen

  2. Introduction • Inductive Logic Programming algorithm: C: Set of Datalog clauses, initially empty D: Database of facts (Knowledge base) repeat make clauses in C more specific (downward refinement) evaluate C against D 2. 1. BNAIC-2003

  3. Database of Facts g1 g2 • {e(g1,n1,n2,a),e(g1,n2,n1,a),e(g1,n2,n3,a), e(g1,n3,n1,b),e(g1,n3,n4,b),e(g1,n3,n5,c), e(g2,n6,n7,b)} n2 n4 n6 a a b b a n1 n5 n7 b c n3 BNAIC-2003

  4. Clause N4 • k(G)  e(G,N1,N2,a),e(G,N2,N3,a), e(G,N1,N4,a),e(G,N4,N5,b) b N5 a N1 a N3 a N2 BNAIC-2003

  5. ={G/g1,N1/n2,N2/n1,N3/n2,N4/n3,N5/n1} Evaluation of a clause • -subsumption: DC iff there is a substitution , (C)  D • Database D: {e(g1,n1,n2,a),e(g1,n2,n1,a),e(g1,n2,n3,a), e(g1,n3,n1,b),e(g1,n3,n4,b),e(g1,n3,n5,c), e(g2,n6,n7,b)} • Clause C: k(G)  e(G,N1,N2,a),e(G,N2,N3,a), e(G,N1,N4,a),e(G,N4,N5,b) BNAIC-2003

  6. a a a b N4 b N1 N5 a a a N1 a N3 a N3 a b N2 N2 N5 N4 Evaluation of a clause g1 g2 n2 n4 n6 a a b b b a n1 n5 n7 b a n3 BNAIC-2003

  7. N3 a N1 Equivalent a N1 N2 a N2 Evaluation of a clause g1 g2 n2 n4 n6 a a b b b a n1 n5 n7 b a n3 N5 N3 BNAIC-2003

  8. N3 a N1 Equivalent a N1 N2 a N2 Evaluation of a clause k(G)  e(G,N1,N2,a) k(G)  e(G,N1,N2,a), e(G,N1,N3,a) n3 N5 N3 BNAIC-2003

  9. N1 N2 a Evaluation of a clause • OI-subsumption: DC iff there is a substitution , (C)  D, while: •  is injective •  does not map to constants in C N3 a N1 a N2 BNAIC-2003

  10. e(G,N1,N2,a) ,e(G,N2,N1,a) ,e(G,N1,N3,a) a Using M only edge labeled trees can be constructed! b a Clause Refinement - modes + old variable- new variable# constant • User defined Refinement using modes [Progol,Aleph,Warmr,Tilde,Farmer] • T={k(G),e(G,N,N,L)} M={e(+,-,-,#),e(+,+,-,#)} • k(G) BNAIC-2003

  11. Clause Refinement - modes • k(G) e(G,N1,N2,a) k(G) e(G,N1,N2,a),e(G,N1,N3,a) • k(G) e(G,N1,N2,a),e(G,N1,N3,a), e(G,N2,N4,a),e(G,N3,N5,b) • Complete & proper refinement is possible with OI-subsumption, not with -subsumption. a b a a BNAIC-2003

  12. Refinement using Primary Keys • Assume we know: between a pair of nodes there is at most one edge with one label • How to incorporate this knowledge in the refinement operator? • M={e(+,-,-,#),e(+,+,-,#),e(+,+,+,#)} • These modes allow: • k(G)  e(G,N1,N2,a), e(G,N1,N2,b) • k(G)  e(G,N1,N2,a), e(G,N1,N2,L1) • Primary key: {1,2,3} (first 3 arguments of e) BNAIC-2003

  13. Expressiveness OI vs -subsumption a ? b • k(G) e(G,N1,N2,a),e(G,N2,N3,L),e(G,N3,N4,b) • For proper and complete refinement:OIis required • Under OI: L  a, L  b BNAIC-2003

  14. In an ideal situation... • We have complete & proper refinement • We are not required to use OI for all types (weak Object Identity) BNAIC-2003

  15. Proper refinement using Primary Keys • In many cases, this ideal situation exists for refinement using primary keys! • k(G) e(G,N1,N2,a),e(G,N2,N3,L),e(G,N3,N4,b) BNAIC-2003

  16. Proper refinement using Primary Keys • T={k(G),e(G,N,N,L),t(L,C)} M={e(+,-,-,-),e(+,+,-,-),t(+,#)}K(p)={1,2,3} K(t)={1,2}OI={G,N} • k(G) e(G,N1,N2,L1),t(L1,a), e(G,N1,N3,L2),t(L2,a) BNAIC-2003

  17. Proper refinement using Primary Keys • Given predicates, types, modes, primary keys and a partition of types into OI and non-OI • We prove refinement is proper and complete if for every mode there is a primary key which does not include any non-OI output. BNAIC-2003

  18. Conclusions • Higher performance for ILP algorithms • primary keys restrict the search space efficiently • refinement is proper • Higher flexibility • weak OI is more flexible than full OI BNAIC-2003

  19. Clause refinement - other representation better? Background knowledge:a(G,N1,N2)  e(G,N1,N2,a)b(G,N1,N2)  e(G,N1,N2,b)e(G,N1,N2)  e(G,N1,N2,L) • k(G) a(G,N1,N2),e(G,N2,N3),b(G,N3,N4) • k(G) a(G,N1,N2),e(G,N1,N2) BNAIC-2003

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